For part(b), we have\begin{equation} (x^2+y^2-1)dx+2xydy=0\end{equation}Note that\begin{equation} M_y=N_x=2y\end{equation}The equation is exact.By integration\begin{equation} H=\frac{1}{3}x^3+xy^2-x+h^\prime(y)\end{equation}\begin{equation} h^\prime(y)=0\end{equation}We choose\begin{equation} h(y)=0\end{equation}In this way,\begin{equation} H(x,y)=\frac{1}{3}x^3+xy^2-x=C\end{equation}I will post solution to other parts later if no one else follows.

Observe that Hessian of $H(x,y)$ is $$\begin{pmatrix} 2x &2y\\2y &2x\end{pmatrix};$$compare with the Jacobi matrix (Jacobian is its determinant). In this particular case (of exact system) sometimes it is called skew-Hessian.

I attach the Contour plot of $H(x,y)$; note that $(-1,0)$ is the local maximum and $(1,0)$ is the local minimum, while $(0,\pm 1)$ are two saddle points